Question

Dry air is passed through a solution containing 10 g of the solute in 90 g of water and then through pure water. The loss in weight of solution is 2.5g and that of pure solvent is 0.05g. The molecular weight of the solute is

Answer 1

Solution.

$$Ps/Po = 2.5 g/(2.5g+0.05g)$$

Ps - partial pressure of solution

$$(Po-Ps)/Ps = 0.05/2.5$$

$$n(H2O) = m(H2O)/M(H2O)$$

n(H2O) = 90/18 = 5.0 mole

$$10/M = n(solution)$$

$$(Po-Ps)/Ps = 0.02$$

$$0.02 = (10/M)/(90/18)$$

M = 100 g/mole

Answer:

M = 100 g/mole